Answer to Question #347481 in Statistics and Probability for glen

Question #347481

what is the mean and variance of samples of three cards are drawn random from a population of seven cards numbered 1 to 7


1
Expert's answer
2022-06-03T06:22:07-0400

We have population values 1,2,3,4,5,6,7 population size N=7 and sample size n=3.

Mean of population (μ)(\mu) = 1+2+3+4+5+6+77=4\dfrac{1+2+3+4+5+6+7}{7}=4

Variance of population 


σ2=Σ(xixˉ)2N=9+4+1+0+1+4=97=4\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=\dfrac{9+4+1+0+1+4=9}{7}=4σ=σ2=21.4142\sigma=\sqrt{\sigma^2}=\sqrt{2}\approx1.4142


The number of possible samples which can be drawn without replacement is NCn=7C3=35.^{N}C_n=^{7}C_3=35.

Mean of sampling distribution 


μXˉ=μ=4\mu_{\bar{X}}=\mu=4


The variance of sampling distribution 


Var(Xˉ)=σXˉ2=σ2n(NnN1)Var(\bar{X})=\sigma^2_{\bar{X}}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})




=43(7371)=89= \dfrac{4}{3}(\dfrac{7-3}{7-1})=\dfrac{8}{9}

The number of possible samples which can be drawn with replacement is Nn=73=343.N^n=7^3=343.

Mean of sampling distribution 


μXˉ=μ=4\mu_{\bar{X}}=\mu=4


The variance of sampling distribution 


Var(Xˉ)=σXˉ2=σ2n=43Var(\bar{X})=\sigma^2_{\bar{X}}= \dfrac{\sigma^2}{n}= \dfrac{4}{3}




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